## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = \frac{{2\left( {{x^2} + 1} \right)}}{{\sqrt {\cos 2x} }}\left( {\frac{{2x}}{{{x^2} + 1}} + \tan 2x} \right)$$
\eqalign{ & y = \frac{{2\left( {{x^2} + 1} \right)}}{{\sqrt {\cos 2x} }} \cr & {\text{Take the natural logarithm on both sides}} \cr & \ln y = \ln \frac{{2\left( {{x^2} + 1} \right)}}{{\sqrt {\cos 2x} }} \cr & {\text{use quotient property for logarithms}} \cr & \ln y = \ln 2\left( {{x^2} + 1} \right) - \ln \sqrt {\cos 2x} \cr & \ln y = \ln \left( {2{x^2} + 2} \right) - \ln {\left( {\cos 2x} \right)^{1/2}} \cr & {\text{use power property for logarithms}} \cr & \ln y = \ln \left( {2{x^2} + 2} \right) - \frac{1}{2}\ln \left( {\cos 2x} \right) \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{4x}}{{2{x^2} + 2}} - \frac{1}{2}\left( {\frac{{ - 2\sin 2x}}{{\cos 2x}}} \right) \cr & {\text{simplifying, we get:}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2} + 1}} + \frac{{\sin 2x}}{{\cos 2x}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2} + 1}} + \tan 2x \cr & {\text{solve the equation for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = y\left( {\frac{{2x}}{{{x^2} + 1}} + \tan 2x} \right) \cr & {\text{replace }}y = \frac{{2\left( {{x^2} + 1} \right)}}{{\sqrt {\cos 2x} }} \cr & \frac{{dy}}{{dx}} = \frac{{2\left( {{x^2} + 1} \right)}}{{\sqrt {\cos 2x} }}\left( {\frac{{2x}}{{{x^2} + 1}} + \tan 2x} \right) \cr}